Question: Simplify the following expression: $y = \dfrac{5x^2+27x+10}{x + 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(10)} &=& 50 \\ {a} + {b} &=& &=& {27} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $50$ and add them together. The factors that add up to ${27}$ will be your ${a}$ and ${b}$ When ${a}$ is ${2}$ and ${b}$ is ${25}$ $ \begin{eqnarray} {ab} &=& ({2})({25}) &=& 50 \\ {a} + {b} &=& {2} + {25} &=& 27 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 +{2}x) + ({25}x +{10}) $ Factor out the common factors: $ x(5x + 2) + 5(5x + 2)$ Now factor out $(5x + 2)$ $ (5x + 2)(x + 5)$ The original expression can therefore be written: $ \dfrac{(5x + 2)(x + 5)}{x + 5}$ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ This leaves us with $5x + 2; x \neq -5$.